参数资料
型号: IR3080MTR
厂商: International Rectifier
文件页数: 30/41页
文件大小: 0K
描述: IC PHASE CONTROLLER 32L MLPQ
标准包装: 3,000
系列: XPhase™
应用: 处理器
电流 - 电源: 11mA
电源电压: 9.5 V ~ 14 V
工作温度: 0°C ~ 100°C
安装类型: 表面贴装
封装/外壳: 32-VFQFN 裸露焊盘
供应商设备封装: 32-MLPQ(5x5)
包装: 带卷 (TR)
IR3080
The phase delay resistor ratios for phases 1 to 6 at 400kHz of switching frequencies are RA PHASE1 =0.628,
RA PHASE2 =0.415, RA PHASE3 =0.202, RA PHASE4 =0.246, RA PHASE5 =0.441 and RA PHASE6 =0.637 starting from down-
slope. Pre-select R PHASE11 =R PHASE21 =R PHASE31 =R PHASE41 =R PHASE51 = R PHASE61 =10k ? ,
R PHASE 12 =
RA PHASE 1
1 ? RA PHASE 1
? R PHASE 11 =
0 . 628
1 ? 0 . 628
? 10 * 10 3 = 16 . 9 k ?
R PHASE22 =7.15k ? , R PHASE32 =2.55k ? , R PHASE42 =3.24k ? , P PHASE52 =7.87k ? , R PHASE62 =17.4k ?
Bootstrap Capacitor C BST
Choose C BST =0.1uF
Decoupling Capacitors for Phase IC and Power Stage
Choose C VCC =0.1uF, C VCCL =0.1uF
VOLTAGE LOOP COMPENSATION
Type II compensation is used for the converter with AL-Polymer output capacitors. Choose the crossover frequency
fc=40kHz, which is 1/10 of the switching frequency per phase, and determine Rcp and C CP .
R CP =
( 2 π ? f C ) 2 ? L E ? C E ? R FB ? V RAMP
V O * 1 + ( 2 π * f C * C * R C ) 2
=
( 2 π ? 40 ? 10 3 ) 2 ? ( 220 ? 10 ? 9 / 6 ) ? ( 560 ? 10 ? 6 ? 10 ) ? 365 ? 0 . 8
( 1 . 35 ? 20 ? 10 ? 3 ) * 1 + ( 2 π * 40 * 10 3 * 560 * 10 ? 6 * 7 * 10 ? 3 ) 2
= 2 . 0 k ?
C CP =
10 ? L E ? C E
R CP
=
10 ? ( 220 ? 10 ? 9 / 6 ) ? ( 560 ? 10 ? 6 * 10 )
2 . 0 ? 10 3
= 71 nF , Choose C CP =68nF
Choose C CP1 =47pF to reduce high frequency noise.
CURRENT SHARE LOOP COMPENSATION
The crossover frequency of the current share loop f CI should be at least one decade lower than that of the voltage
loop f C . Choose the crossover frequency of current share loop f CI =4kHz , and calculate C SCOMP ,
F MI =
R PWMRMP * C PWMRMP * f SW * V PWMRMP
( V I ? V PWMRMP ? V DAC ) * ( V I ? V DAC )
=
16 . 2 * 10 3 * 220 * 10 ? 12 * 400 * 10 3 * 0 . 8
( 12 ? 0 . 8 ? 1 . 35 ) * ( 12 ? 1 . 35 )
= 0 . 011
C SCOMP =
=
0 .65 * R PWMRMP * V I * I O * G CS _ ROOM * R LE * [1 + 2 π * f CI * C E * ( V O I O )] * F MI
V O ? 2 π ? f CI * 1 . 05 * 10 6
0. 65 * 16 . 2 * 10 3 * 12 * 105 * 34 * ( 0 .47 * 10 ? 3 6) * [1 + 2 π * 4 * 10 3 * 560 * 10 ? 6 * 10 * (1. 33 ? 105 * 9. 1 * 10 ? 4 ) 105] * 0. 011
( 1 . 33 ? 105 * 9 . 1 * 10 ? 4 ) ? 2 π ? 4 * 10 3 * 1 . 05 * 10 6
= 31 . 4 nF
Choose C SCOMP =33nF.
Page 30 of 41
9 /30/04
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